Instead of finding the longest common
subsequence, let us try to determine the
length of the LCS.
Then tracking back to find the LCS.
Consider a1a2…am and b1b2…bn.
Case 1: am=bn. The LCS must contain am,
we have to find the LCS of a1a2…am-1 and
b1b2…bn-1.
Case 2: am≠bn. Wehave to find the LCS of
a1a2…am-1 and b1b2…bn, and a1a2…am and
b b b
b1b2…bn-1
Let A = a1 a2 … am and B = b1 b2 … bn
Let Li j denote the length of the longest i,g g
common subsequence of a1 a2 … ai and b1 b2
… bj.
Li,j = Li-1,j-1 + 1 if ai=bj
max{ L L } a≠b i-1,j, i,j-1 if ai≠j
L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
標簽：
the
subsequence
determine
Instead
上傳時間：
2013-12-17
上傳用戶：evil
//初始化
initscr()
//獲得屏幕尺寸
getmaxyx(stdscr, h, w)
//畫背景
for(i=0 i<h i++)
for(j=0 j<w j++){
mvaddch(i, j, ACS_CKBOARD)
}
refresh()
//建立窗口
pad = newpad(80, 128)
for(i=0 i<80 i++){
char line[128]
sprintf(line, "This line in pad is numbered d\n", i)
mvwprintw(pad, i, 0, line)
}
//刷新屏幕
refresh()
prefresh(pad, 0, 1, 5, 10, 20, 45)
for(i=0 i<50 i++){
prefresh(pad, i+1, 1, 5, 10, 20, 45)
usleep(30000)
}
//等待按鍵
getch()
標簽：
getmaxyx
initscr
stdscr
for
上傳時間：
2014-08-30
上傳用戶：龍飛艇
